why does Zn become oxidised in the first place in operation of voltaic cell?

I am so confused that during the operation of a voltaic cell, why does $\ce{Zn}$ gets oxidized? In anode side, there is a zinc metal in $\ce{ZnSO4}$ solution, but why does the $\ce{Zn}$ lost 2 electrons?

• I think you need to clarify this question further. Which voltaic cell are you talking about? The reason Zn oxidizes is very coupled to what is reducing on the other side. The reaction is not producing free electrons, those electrons are reducing something else. – Burak Ulgut Jul 19 '16 at 14:20

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. In anode compartment you have Zn in contact with ZnSO4. In cathode cell you have two other reagents. For example Cu and CuSO4. activity series of the metals suggest that zinc is OK to lose electrons and turn into zinc ion. Copper ions from CuSO4 are strongly willing to turn into reduced copper metal. So zinc gives away electrons to copper ions because Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$.
Why Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$ is a separate question and it belongs to a different field (quantum mechanics/molecular orbital theory rather than electrochemistry).
In short - there is Zn metal that holds electrons weakly and Cu$^{2+}$ that will hold electrons strongly when it forms Cu metal. Hence, electrons flow to Cu$^{2+}$ from Zn.